Question

two samples are taken with the following sample means, standard deviations, and sample sizes.
$\bar{x}_1 = 28$
$s_1 = 4$
$n_1 = 55$

$\bar{x}_2 = 32$
$s_2 = 2$
$n_2 = 51$

estimate the difference in population means using a 99% confidence level. use a calculator.
round answers to 1 decimal place.

Explanation:

Step1: Identify the formula for confidence - interval for the difference in means

The formula for the confidence - interval for the difference in two population means $\mu_1-\mu_2$ when the population standard deviations are unknown is $(\bar{x}_1 - \bar{x}_2)\pm t_{\alpha/2}\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}$. The degrees of freedom is calculated using the formula $df=\min(n_1 - 1,n_2 - 1)=\min(55 - 1,51 - 1)=50$. For a 99% confidence level, $\alpha=1 - 0.99 = 0.01$ and $\alpha/2=0.005$. Looking up in the t - distribution table, $t_{0.005,50}\approx 2.678$.

Step2: Calculate the point estimate

The point estimate of $\mu_1-\mu_2$ is $\bar{x}_1-\bar{x}_2=28 - 32=-4$.

Step3: Calculate the margin of error

$E=t_{\alpha/2}\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}=2.678\sqrt{\frac{4^{2}}{55}+\frac{2^{2}}{51}}$. First, calculate $\frac{4^{2}}{55}+\frac{2^{2}}{51}=\frac{16}{55}+\frac{4}{51}=\frac{16\times51 + 4\times55}{55\times51}=\frac{816+220}{2805}=\frac{1036}{2805}\approx0.369$. Then $\sqrt{\frac{1036}{2805}}\approx0.607$. And $E = 2.678\times0.607\approx1.625$.

Step4: Calculate the confidence interval

The confidence interval is $(\bar{x}_1-\bar{x}_2)-E<\mu_1 - \mu_2<(\bar{x}_1-\bar{x}_2)+E$. Substituting the values, we get $-4-1.625<\mu_1 - \mu_2<-4 + 1.625$, which simplifies to $-5.6<\mu_1 - \mu_2<-2.4$.

Answer:

$(-5.6,-2.4)$